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I will answer this tenatively , as no one has yet.....

To get from circle one to circle two, you will need to

Translate 10 to the left  (8-(-2))    AND

Translate DOWN 4   (5-1)    AND

introduce a dilation factor of 1/3     (1/3 x 6 =2)    

(x+6) /(sqrt(x^2-3x-4) )

Has a domain for ALL x such that the denominator does NOT equal Zero..AND we do not have a NEGATIVE sqrt

let's find what those values are

sqrt(x^2-3x-4)  = sqrt(x-4)(x+1)       This WILL equal zero when x = 4 or -1 <------ NOT allowed !

ALSO from    - infinity to +4 is invalid as a negative sqrt will result

to NET it all out , the domain is         4< x  thru  + infinity

If you're buying a pie that is 20% OFF...you wind up paying 80% of the ORIGINAL price

$18 x 80% = 18 x .8  = $ 14.40     

You would SAVE 20% of the original price

$ 18  x  20%  = 18 x 0.2 = $ 3.60

Just COUNT them...as written this is an EXPRESSION with TWO terms.    'u'    and   '5s'

A ONE MOLAR solution has ONE MOLE dissolved in 1 liter of solution.

Silver nitrate wieighs   107.868 + 14.007 + (3)15.999 = 169.872 gm/mole

169.872 gm/mole x .25 mole/liter x 500/1000 l  =  21.234 gm

just to talk

for what?

yeppers!

Sounds OK to me.....

 ....or you could just do it this way...Building on CPhil's answer

4s=s^2     divide by s

4=s

2.

Position the lower left corner, A , at (0,0)

The equation of AD  is   y  = x

And  let  B  be  positioned at (4,0)   

And the equation of  BC    is   y = (-5/4) ( x  - 4)  =  (-5/4)x + 5

Setting  these two equations equal and solving for x we have

x  = (-5/4)x + 5

(9/4) x  = 5

x  = 20/9  =   y  =  height of triangle ABE

So....the area of ABE  = (1/2) (base) (height)  = (1/2) (4) (20/9)  =

40/9  units^2  

4:  The number of inches in the perimeter of a square is equal to the number of square inches in its area. Find the length, in inches, of a side of the square.

Let S  = the side of the square

P  = 4S

A  = S^2

This impies that

4S  = S^2

S^2  - 4S  = 0     factor

S ( S - 4)  = 0

Setting both factors to 0  and solving for S  produces  S  = 0  and S  =4

Reject the first....accept the second

S  = 4 in

Looking ofr equation of a line of the form    y = mx+b   where m = slope and b=y axis intercept

y = 2/5 x + b          Now substitue in the point given (-10,3) to find 'b'

3 = 2/5 (-10) + b

3 = -4 + b

 b = 7

Equation becomes:  y = 2/5  x + 7

1.Let the side  of the smaller triangle   = S

Since the area  of the larger triangle is 16 times as much....each dimension must be 4 times that of the smaller triangle...so the length of its side must be  4S

And the sum of ther perimeters = 45....so

3S  + 3(4S)  = 45

15S  = 45

S  = 3 in

So....the side of the larger trinagle  = 4S  = 4(3)  =12 in

And its area  =  (1/2) 12^2 sin(60)  = 72√3 / 2   =    36√3  in^2

((sqrt(3))^2 - 2 (sqrt(2))^2)/(sqrt(3) - 2 sqrt(2))=z

Multiply exponents. (sqrt(2))^2 = 2^(2/2):

((sqrt(3))^2 - 22^(2/2))/(sqrt(3) - 2 sqrt(2))

((sqrt(3))^2 - 2×2)/(sqrt(3) - 2 sqrt(2))

Cancel exponents. (sqrt(3))^2 = 3:

(3 - 2×2)/(sqrt(3) - 2 sqrt(2))

(-4 + 3)/(sqrt(3) - 2 sqrt(2))

(-1)/(sqrt(3) - 2 sqrt(2))

Multiply numerator and denominator of (-1)/(sqrt(3) - 2 sqrt(2)) by -1:

1/(2 sqrt(2) - sqrt(3))

Multiply numerator and denominator of 1/(2 sqrt(2) - sqrt(3)) by 2 sqrt(2) + sqrt(3):

(2 sqrt(2) + sqrt(3))/((2 sqrt(2) - sqrt(3)) (2 sqrt(2) + sqrt(3)))

(2 sqrt(2) - sqrt(3)) (2 sqrt(2) + sqrt(3)) = 2 sqrt(2)×2 sqrt(2) + 2 sqrt(2) sqrt(3) - sqrt(3)×2 sqrt(2) - sqrt(3) sqrt(3) = 8 + 2 sqrt(6) - 2 sqrt(6) - 3 = 5:

z = (2sqrt(2) + sqrt(3)) / 5

g(x)   = y

y  =       3                    2 (x - 4)

        _______    +      _________

        x^2 - 16             (x + 4)(x - 4)

y  =    3 + 2x - 8

        __________

          x^2  - 16

y  =  2x -  5

       _______

       x^2  - 16

y (x^2 - 16)  = 2x - 5

yx^2 - 16y  = 2x - 5

16y  =  yx^2 - 2x + 5

16y  = y (x^2 - (2/y)x + 5/y)

16 =  x^2  - (2/y)x + 5/y      complete the square on x

16  = x^2 - (2/y)x + 5/y + 1/y^2 - 1/y^2 

Add  1/y^2  to both sides....subtract 5/y from both sides

16 + 1/y^2 -5/y   = x^2 -(2/y)x + 1/y^2      factor the right side

16 + 1/y^2 - 5/y  = (x - 1/y)^2

1/y^2 - 5/y + 16  =  (x - 1/y)^2      get a common denominator on the left

[16y^2 -5y + 1] / y^2  =  (x - 1/y)^2      take both roots

±√[ (16y^2 -5y + 1) / y^2 ]  =  x - 1/y      add 1/y to both sides

1/y ±√  (16y^2 -5y + 1) / y

[ 1 ±√  (16y^2 -5y + 1)  ] / y  = x       "swap" x and y

[ 1 ±√ [ 16x^2 - 5x + 1) ] / x  =  y  =  g^-1(x)

Thus....this is the inverse.....but it is not one-to-one.....we have two values for g^-1(5)

g^-1 (5)  =   ( 1 + √ [ 16(5)^2 - 5(5) + 1 ] )  / 5   = ( 1 + √ 376 ) /  5  ≈ 4.078

g^-1(5) =  ( 1 - √ [ 16(5)^2 - 5(5) + 1 ] )  / 5   = ( 1 - √ 376 ) /  5  ≈ -3.678

So   g-1 (5)   =   ≈4.078    and   ≈ -3.678

I am still confused.

I still don’t get it.

The question states the drip fills the container at  2 in/hr

    since the variable is 't'  ...as in hours....if you multiply   hrs x 2 in/hr    the 'hr' will cancel and you will be left with inches...

the coefficient should be   '2'

Just solve the given equation for 'x'

100-2x = 84       subtract 100 from both sides of the equation

-2x = -16            Divide both sides by  - 2

x = 8

He missed 8 questions      

Thanks, heureka....here's my approach....

Using the Law of Cosines we have that

12^2 = (2CD)^2 + 6^2  -  [ 2 * 2CD * 6 ] cos BDA

144 = 4CD^2  + 36 - 24CDcosBDA

108 - 4CD^2  = - 24CDcosBDA

[ 4CD^2 - 108 ] / 24CD  = cos BDA

[CD^2 - 27 ] / 6CD  =cos BDA

Since BDA  and CDA  are supplementary.....-cos BDA  =  cosCDA

Using it again and substituting, we have

9^2 = CD^2 + 6^2  - [ 2 * CD * 6 ] [-cos BDA]

81  = CD^2 + 36 + 12CD  [ CD^2 - 27] / 6CD ]

45 = CD^2 + 2[CD^2 - 27 ]

45 = CD^2 + 2CD^2 - 54

99 = 3CD^2

33 = CD^2

√33 = CD

And  BD  is twice this

So

BD + CD  =  BC  =  √33  + 2√33  =   3√33

Here's a pic :